https://www.programmercarl.com/kamacoder/0101.%E5%AD%A4%E5%B2%9B%E7%9A%84%E6%80%BB%E9%9D%A2%E7%A7%AF.html

1.本题我以为是统计所以陆地的面积，结果是孤岛也就是不挨着边界的陆地总面积，说起来我觉得并不简单，想不到将与边界相连接的陆地变成水，本题要求找到不靠边的陆地面积，那么我们只要从周边找到陆地然后 通过 dfs或者bfs 将周边靠陆地且相邻的陆地都变成海洋，然后再去重新遍历地图 统计此时还剩下的陆地就可以了。

CPP

#include<bits/stdc++.h>
using namespace std;
int cnt;
int dir[4][2] = {1,0,0,1,0,-1,-1,0};

void dfs(vector<vector<int>>& grid, int x, int y){
    grid[x][y] = 0;
    cnt++;
    
    for(int i = 0; i < 4; i++){
        
        int nextx = x + dir[i][0];
        int nexty = y + dir[i][1];
        if(nextx < 0 || nextx >= grid.size() || nexty < 0 || nexty >= grid[0].size())continue;
        if(grid[nextx][nexty] == 0)continue;
        dfs(grid, nextx, nexty);
        
    }
    return ;
}

int main(){
    
    int n, m;
    cin >> n >> m;
    
    vector<vector<int>> grid(n, vector<int>(m, 0));
    
    for(int i = 0; i < n; i++){
        for(int j = 0; j < m; j++){
            cin >> grid[i][j];
        }
    }
    
    for(int i = 0; i < n; i++){
        if(grid[i][0] == 1)dfs(grid, i, 0);
        if(grid[i][m - 1] == 1)dfs(grid, i, m - 1);
    }
    
    for(int i = 0; i < m; i++){
        if(grid[0][i] == 1)dfs(grid, 0, i);
        if(grid[n - 1][i] == 1)dfs(grid, n - 1, i);
    }
    cnt = 0;
    for(int i = 0; i < n; i++){
        for(int j = 0; j < m; j++){
            if(grid[i][j] == 1){
                dfs(grid, i, j);
            }
        }
    }
    cout << cnt << endl;
    
    
}